Question: You have found the following ages (in years) of all $4$ zebras at your local zoo: $ 14,\enspace 15,\enspace 9,\enspace 1$ What is the average age of the zebras at your zoo? What is the variance? Round your answers to the nearest tenth. Average age: $ $
Answer: Because we have data for all $4$ zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$. To find the population mean, add up the values of all $4$ ages and divide by $4$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{14 + 15 + 9 + 1}{{4}} = {9.75\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $14$ years $4.25$ years $18.06$ years $^2$ $15$ years $5.25$ years $27.56$ years $^2$ $9$ years $-0.75$ years $0.56$ years $^2$ $1$ year $-8.75$ years $76.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the population variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{18.06} + {27.56} + {0.56} + {76.56}} {{4}} $ $ {\sigma^2} = \dfrac{{122.74}}{{4}} = {30.69\text{ years}^2} $ The average zebra at the zoo is $9.8$ years old. The population variance is $30.7$ years $^2$.